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  #21  
Old 08-25-2012, 07:13 AM
aeroscott aeroscott is offline
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Default Re: AC induction motor rewinding questions

good find , I would think this would apply to switched reluctance motors as well .
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  #22  
Old 08-25-2012, 07:38 AM
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Default Re: AC induction motor rewinding questions

Sorry, but I just could not resist replying.

Quote:
But AFAIK, there shouldn't be any inherent mechanism why lower voltage would be so much more demanding?
It is called the "free lunch" principal. There is no such thing as a "free lunch".

You might be a little young to have seen this, but:

In 1952, Ford was in the middle of changing over from a 6volt DC electrical system to a newer 12Volt DC system.

The 6volt wiring was of Huge gauge and so it handled the 12volt change over well.

The bulbs were changed to 12VDC as was the generator and starter.

The first thing you noticed when the two starters were side by side on the bench, was the huge gauge windings on the 6VDC starter. The 12VDC unit was much smaller...almost exactly half as large, with twice the windings.

12 volts-1/2 wire size-twice the windings= same work performed.
6Volts-double the wire size-half the windings= same work performed.

There is a close relationship between voltage and amperage (current).

Lower voltage requires a larger amperage/current to get the same work as a higher voltage. Thus needing larger conductors to handle the increased current.

http://www.allaboutcircuits.com/vol_1/chpt_2/1.html
http://www.sengpielaudio.com/calculator-ohm.htm
http://jersey.uoregon.edu/vlab/Voltage/
http://www.the12volt.com/ohm/ohmslaw.asp

A lot of reading, I know, but worth it, Miz
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  #23  
Old 08-25-2012, 08:19 AM
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Default Re: AC induction motor rewinding questions

If I try once again;

In A MOTOR, there is a specific amount of SPACE for windings IN THE SLOTS. They are filled with COPPER WIRE. For lower voltage design, there are FEWER TURNS and hence thicker wire and/or more wires can be fitted to the same slot space, for MORE CURRENT. Hence, there is no direct relationship between motor voltage and efficiency or power; instead, these are "minor" optimization things. So in practice, you can decide whether you design a low-voltage high-current motor or a high-voltage low-current motor; within certain practical limits, with a certain sweet spot.

A very good example would be a series wound DC motor using as low voltages as 100V to produce power as high as 50 kW. It is only slightly suboptimal compared to a higher-voltage system, but doable, and many DIY conversions use these. Let me say again that WE are going for a MID-VOLTAGE system, so actually from an EV viewpoint, the current-voltage-power ratio should be quite well balanced near the sweet spot.

This has nothing to do with free lunch. It is just about paying your lunch in U.S. dollars -- you need a few of them -- or in yen -- you need a lot of them, but both have the same value; you can get a lunch for $20 or 2000 yen, and you can get 10 kilowatts with 100V and 100A, or 300V and 33A, or 1000V and 10A.

As I stated above, generally a higher voltage is somewhat better.

Last edited by Siwastaja; 08-25-2012 at 08:52 AM. Reason: Removed unnecessary ranting and egoism, sorry. Read http://bikeshed.org/ instead!
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  #24  
Old 08-25-2012, 08:43 AM
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Default Re: AC induction motor rewinding questions

Wow, I'm not an EE, and I don't rewind ac motors, but I am learning all I can. So is everyone here. This thread is the path I want to take if/when I can start an EV project. Thanks for the info!
I am, however, an accomplished Internet Troll.
That, and its Saturday morn, and I'm bored.

Last edited by few2many; 08-25-2012 at 09:05 AM. Reason: Removed unneccessary...uh, calling out/trolling?
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  #25  
Old 08-25-2012, 09:07 AM
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Default Re: AC induction motor rewinding questions

Quote:
Originally Posted by PStechPaul View Post
It takes about 18kW to maintain reasonable speed on the highway, assuming 300Wh per mile and 60 MPH. At 240 VAC three phase that's about 43 amps per phase. #17 AWG with 90C insulation is good for about 16 amps. So three conductors should be OK.
http://www.armstrongssupply.com/wire_chart.htm

I could not find a good motor winding wire size chart.

Mag wire data

http://www.coilwinder.com/Magnet%20wire%20data.htm

So 43 amps is the max ever? come on, we no better PS

What is the max amps the controller puts out. Can't believe you guys.
I give up..
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  #26  
Old 08-25-2012, 09:32 AM
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Default Re: AC induction motor rewinding questions

Looks more like he is considering Cruising volts/amps, not max. Huge difference.
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  #27  
Old 08-25-2012, 01:35 PM
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Thumbs up Re: AC induction motor rewinding questions

Quote:
Originally Posted by few2many View Post
Looks more like he is considering Cruising volts/amps, not max. Huge difference.
Yes, of course. Most of the time, for normal driving (not drag racing or hill climbing) an average size commuter/family type EV will use an average of about 200-350 Wh per mile. I just used a nominal value and 60 MPH for a ballpark estimate. Actually, using my EVcalculator, a 1000 kg vehicle at 100 km/h requires only about 5.5 kW to maintain speed on a level road, and about 19 kW for a 5% grade or accelerating at 0.05 G (1 MPH/sec) at that speed. 15 kW will also provide acceleration of 2 m/s/s at an average speed of 25 km/h, but we are considering a motor and controller that can provide about 3x torque (current) for short bursts, so at 45 kW and average speed of 50 km/h you can accelerate at 3 m/s/s or 6.7 MPH/s which is 0-60 in under 10 seconds.

Wire will withstand a 3x current overload for 10 seconds with no problem as long as it's allowed to cool for a few minutes. This is why circuit breakers and motor overloads have time delay curves, many of which allow 90 seconds at 3x current and do not trip instantaneously until 8x-15x rated current.

The wire chart provided by coilwinder.com is helpful but what I'm looking for is the proper size for winding a motor based on current. I realize that this depends on the I^2R losses and the thermal characteristics of the motor and ambient temperature and cooling methods and many other factors, but surely there is a "rule of thumb" for ACIM stator windings based on current and insulation temperature class. The closest I could find was in these FAQs: http://www.mwswire.com/faqs.htm#mw6

Quote:
Q: How do I calculate the amperage for a given round magnet wire?
Quote:
A: The formula for current carrying capacity in amperes for copper magnet wire wound into a coil is: d2 x 4869.48 (d = diameter in inches). This formula is pretty conservative, and formulas from other sources based on straight lengths of solid or stranded conductors in ambient air may indicate greater current carrying capacity than this one.

Last edited by PStechPaul; 08-25-2012 at 01:51 PM.
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  #28  
Old 08-25-2012, 04:22 PM
aeroscott aeroscott is offline
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Default Re: AC induction motor rewinding questions

Quote:
Originally Posted by mizlplix View Post
Sorry, but I just could not resist replying.



It is called the "free lunch" principal. There is no such thing as a "free lunch".

You might be a little young to have seen this, but:

In 1952, Ford was in the middle of changing over from a 6volt DC electrical system to a newer 12Volt DC system.

The 6volt wiring was of Huge gauge and so it handled the 12volt change over well.

The bulbs were changed to 12VDC as was the generator and starter.

The first thing you noticed when the two starters were side by side on the bench, was the huge gauge windings on the 6VDC starter. The 12VDC unit was much smaller...almost exactly half as large, with twice the windings.

12 volts-1/2 wire size-twice the windings= same work performed.
6Volts-double the wire size-half the windings= same work performed.

There is a close relationship between voltage and amperage (current).

Lower voltage requires a larger amperage/current to get the same work as a higher voltage. Thus needing larger conductors to handle the increased current.

http://www.allaboutcircuits.com/vol_1/chpt_2/1.html
http://www.sengpielaudio.com/calculator-ohm.htm
http://jersey.uoregon.edu/vlab/Voltage/
http://www.the12volt.com/ohm/ohmslaw.asp

A lot of reading, I know, but worth it, Miz
same is true for 12v to 24v , I can start a engine on a bad connection , week batteries and have higher motor efficiency . But in the case of higher rpm's with high number of turns the back emf goes beyond what our battery packs can do(counteract) . If we take that 6v starter added more gear reduction ( for higher starter speed ) and a motor controller . Say we have 24 v pack limit . we can get more speed at higher amps out of the 6v motor then the 12v motor. Some racing dc motors are lower voltage for the same reason ( packs >300v.)

Last edited by aeroscott; 08-25-2012 at 04:24 PM. Reason: added ()
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  #29  
Old 08-25-2012, 05:42 PM
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Ivansgarage Ivansgarage is offline
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Default Re: AC induction motor rewinding questions

Quote:
Originally Posted by PStechPaul View Post

Wire will withstand a 3x current overload for 10 seconds with no problem as long as it's allowed to cool for a few minutes. This is why circuit breakers and motor overloads have time delay curves, many of which allow 90 seconds at 3x current and do not trip instantaneously until 8x-15x rated current.

The wire chart provided by coilwinder.com is helpful but what I'm looking for is the proper size for winding a motor based on current. I realize that this depends on the I^2R losses and the thermal characteristics of the motor and ambient temperature and cooling methods and many other factors, but surely there is a "rule of thumb" for ACIM stator windings based on current and insulation temperature class. The closest I could find was in these FAQs: http://www.mwswire.com/faqs.htm#mw6

[B]
Wouldn't 3 times the current be locked rotor amps? 90 seconds?
no fricken way.

Rule: 300 the lowest to 500 circular mils can handle ONE amp. Most motor
shops will use 500 to 700 circular mils per amp.

We are talking mag wire.........

3 #17 gauge wires = 6156 circular mils divide by 300 the absolute lowest, = 20.5 amps
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Last edited by Ivansgarage; 08-25-2012 at 05:48 PM.
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  #30  
Old 08-25-2012, 06:21 PM
Dennis Dennis is offline
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Default Re: AC induction motor rewinding questions

I don't know if this will help you, Siwastaja, but my little black book that is very informative has some information that might be of some use to you. I'll quote some passages from it.


Changing Concentric windings to Lap Windings

The first method that we shall explain works best if all the coils in the concentric winding have the same number of turns. This method works with six-group consequent-pole windings and also with 12-group concentric windings that have one coil per slot. The following is a description of this method.

Example: A 36-slot, four pole concentric winding with 40 turns per coil: 40 turns / 1.9 = 21 turns per coil. Most lap-wound, three-phase motors are spanned at 80 percent of full span. The formula for this is (slots/poles) + 1*0.8= (36/4) + 1*.8 = 8. Thus, it is a 1 and 8 span. A four-pole lap winding has 12 groups. Coils per group = slots/groups, 0r 36/12 = 3 coils per group. The data for the new lap winding are 21 turns per coil, span 1-8, 12 groups of three coils. The wire size and connection remain the same.


When the coils do not have the same number of turns, the best method to convert concentric to lap is to find the number of effective turns of the concentric winding and then to design a lap winding with the same number of effective turns. To find the number of effective turns, the chord factor must be used. Chord factor tables do not always have enough data for all situations. The following explanation will enable the repairman to understand how these data have been obtained.

Chord Factor

The chord factor is a multiplier used to find the number of effective turns in a coil of wire. The components in determining chord factor are (1) the number of teeth in the stator, (2) the number of poles in the stator, and (3) the span or pitch of the coil, which will determine the number of teeth surrounded by the coil.


Each tooth in a stator represents a number of electrical degrees. The formula to determine the number of electrical degrees per tooth is
(180 degrees* poles) / number of teeth in the stator = degrees per tooth. The degrees per tooth encompassed by a coil are added to get the angle of the coil. The sine of 1/2 of this angle is the chord factor.

A coil with a small span will have a low chord factor and have fewer effective turns than will a coil with a full span. A coil that is over a full span, for example, a 1-11 span (four-pole, 36 slot stator) will have the same chord factor as will a coil with a 1-9 span. Concentric windings will often be over full span.


The following shows how to convert to a lap winding with a concentric-wound, four-pole, 36-slot motor with a different number of turns in each coil. The original winding data are converted to effective turns.

Code:
Span    Turns        Chord factor            Effective turns
1-9        50      *      .984               =     49.2
1-7        32      *      .866               =     27.7
1-5        12      *      .642               =      7.7
                                                         ____
                                                          84.6 Total effective turns per group
The next step is to determine the number of coils per group for the lap winding: slots/groups = 36 slots /12 groups = 3 coils per group. A 1-8 span will be used as in the previous example, using the formula (slots/poles) + 1*0.8 = span. The chord factor for a 1-8 span is 0.939. The effective turns of one group from the old winding are 84.6 turns. The number of new winding 84.6/0.939 = 90 turns per group in the new winding, span 1-8. The connection and wire size remain the same.


Electric Motor Repair 3rd. Edition
Robert Rosenberg and August hand
ISBN 0-03-059584-3

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